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The locus of \(z\) satisfying \(\left|\frac{z-i}{z-2 i}\right|=2\) is a
Options:
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Verified Answer
The correct answer is:
Circle
It is given that,
\(\left|\frac{z-i}{z-2 i}\right|=2\)
Let \(z=x+i y\), then
\(\begin{aligned}
& |x+i(y-1)|=2|x+(y-2) i| \\
\Rightarrow \quad & x^2+(y-1)^2=4\left[x^2+(y-2)^2\right] \\
\Rightarrow & 3 x^2+3 y^2-14 y+16=0 \text { represent a circle, so }
\end{aligned}\)
locus of \(z\) is a circle.
Hence, option (b) is correct.
\(\left|\frac{z-i}{z-2 i}\right|=2\)
Let \(z=x+i y\), then
\(\begin{aligned}
& |x+i(y-1)|=2|x+(y-2) i| \\
\Rightarrow \quad & x^2+(y-1)^2=4\left[x^2+(y-2)^2\right] \\
\Rightarrow & 3 x^2+3 y^2-14 y+16=0 \text { represent a circle, so }
\end{aligned}\)
locus of \(z\) is a circle.
Hence, option (b) is correct.
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